Answer :
To determine which statement accurately describes the situation, let's analyze the behavior of both functions [tex]\( v(n) \)[/tex] and [tex]\( s(n) \)[/tex] as [tex]\( n \)[/tex] (the number of days since the exhibits opened) increases.
### Visual Arts Exhibit
The function for the number of daily visitors at the Visual Arts Exhibit is given by:
[tex]\[ v(n) = \frac{50}{a} + 120 \][/tex]
Notice that in this equation, the term [tex]\(\frac{50}{a}\)[/tex] does not directly depend on [tex]\( n \)[/tex]. Given that [tex]\( a \)[/tex] might be related to [tex]\( n \)[/tex] in practice, let's consider [tex]\( a = n \)[/tex] for understanding the behavior as [tex]\( n \)[/tex] increases.
When [tex]\( a = n \)[/tex]:
[tex]\[ v(n) = \frac{50}{n} + 120 \][/tex]
As [tex]\( n \)[/tex] increases, the term [tex]\(\frac{50}{n}\)[/tex] becomes very small because the denominator gets larger. Thus, [tex]\( \frac{50}{n} \)[/tex] approaches 0. Therefore, [tex]\( v(n) \)[/tex] approaches:
[tex]\[ v(n) \approx 120 \][/tex]
### Sculpture Exhibit
We are given the following values:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline n & 10 & 50 & 100 & 120 & 200 \\ \hline s(n) & 210 & 154 & 147 & 145 & 143 \\ \hline \end{array} \][/tex]
By examining the values in the table, we can observe that as [tex]\( n \)[/tex] increases, [tex]\( s(n) \)[/tex] decreases. It seems to decrease progressively slower and appears to level off. For large values of [tex]\( n \)[/tex], [tex]\( s(n) \)[/tex] seems to approach a value around 143.
### Conclusion
Comparing the behaviors of both functions as [tex]\( n \)[/tex] increases:
- [tex]\( v(n) \)[/tex] levels off to 120.
- [tex]\( s(n) \)[/tex] levels off to around 143.
Thus, as the number of days increases, the number of daily visitors at the Visual Arts Exhibit levels off to 120, which is a lower amount compared to the daily visitors at the Sculpture Exhibit leveling off to around 143.
Therefore, the correct answer is:
A. As the number of days increases, the number of daily visitors at the visual arts exhibit levels off to a lower amount than the number of daily visitors at the sculpture exhibit.
### Visual Arts Exhibit
The function for the number of daily visitors at the Visual Arts Exhibit is given by:
[tex]\[ v(n) = \frac{50}{a} + 120 \][/tex]
Notice that in this equation, the term [tex]\(\frac{50}{a}\)[/tex] does not directly depend on [tex]\( n \)[/tex]. Given that [tex]\( a \)[/tex] might be related to [tex]\( n \)[/tex] in practice, let's consider [tex]\( a = n \)[/tex] for understanding the behavior as [tex]\( n \)[/tex] increases.
When [tex]\( a = n \)[/tex]:
[tex]\[ v(n) = \frac{50}{n} + 120 \][/tex]
As [tex]\( n \)[/tex] increases, the term [tex]\(\frac{50}{n}\)[/tex] becomes very small because the denominator gets larger. Thus, [tex]\( \frac{50}{n} \)[/tex] approaches 0. Therefore, [tex]\( v(n) \)[/tex] approaches:
[tex]\[ v(n) \approx 120 \][/tex]
### Sculpture Exhibit
We are given the following values:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline n & 10 & 50 & 100 & 120 & 200 \\ \hline s(n) & 210 & 154 & 147 & 145 & 143 \\ \hline \end{array} \][/tex]
By examining the values in the table, we can observe that as [tex]\( n \)[/tex] increases, [tex]\( s(n) \)[/tex] decreases. It seems to decrease progressively slower and appears to level off. For large values of [tex]\( n \)[/tex], [tex]\( s(n) \)[/tex] seems to approach a value around 143.
### Conclusion
Comparing the behaviors of both functions as [tex]\( n \)[/tex] increases:
- [tex]\( v(n) \)[/tex] levels off to 120.
- [tex]\( s(n) \)[/tex] levels off to around 143.
Thus, as the number of days increases, the number of daily visitors at the Visual Arts Exhibit levels off to 120, which is a lower amount compared to the daily visitors at the Sculpture Exhibit leveling off to around 143.
Therefore, the correct answer is:
A. As the number of days increases, the number of daily visitors at the visual arts exhibit levels off to a lower amount than the number of daily visitors at the sculpture exhibit.